A022025 Define the sequence S(a(0),a(1)) by a(n+2) is the least integer such that a(n+2)/a(n+1) > a(n+1)/a(n) for n >= 0. This is S(6,102).
6, 102, 1735, 29513, 502028, 8539699, 145263729, 2470994700, 42032617843, 714991805825, 12162299391068, 206885624804179, 3519208035780561, 59863150041598764, 1018296359995701043, 17321632357467588641, 294647962336362325244, 5012080843035687303187
Offset: 0
Keywords
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..811
- D. W. Boyd, Linear recurrence relations for some generalized Pisot sequences, Advances in Number Theory ( Kingston ON, 1991) 333-340, Oxford Sci. Publ., Oxford Univ. Press, New York, 1993.
- Index entries for Pisot sequences
Programs
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Maple
a:= proc(n) option remember; `if`(n<2, [6, 102][n+1], floor(a(n-1)^2/a(n-2))+1) end: seq(a(n), n=0..20); # Alois P. Heinz, Sep 18 2015 -
Mathematica
a[n_] := a[n] = Switch[n, 0, 6, 1, 102, _, 1 + Floor[a[n-1]^2/a[n-2]]]; a /@ Range[0, 20] (* Jean-François Alcover, Nov 16 2020, after Alois P. Heinz *)
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PARI
a=[6,102];for(n=2,30,a=concat(a,a[n]^2\a[n-1]+1));a \\ M. F. Hasler, Feb 10 2016
Formula
The conjectured g.f. (6-5*x^2)/(1-17*x-x^2+14*x^3) yields the same initial terms a(0..271) but from a(272) on a different sequence. - Bruno Berselli and M. F. Hasler, Feb 11 2016
a(n+1) = floor(a(n)^2/a(n-1))+1 for all n > 0. - M. F. Hasler, Feb 10 2016
Extensions
An incorrect program was removed by Alois P. Heinz, Apr 27 2019